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Chapter 5: Problem 94

A compound contains only carbon, hydrogen, and oxygen. Combustion of \(10.68 \mathrm{mg}\) of the compound yields \(16.01 \mathrm{mg}\) \(\mathrm{CO}_{2}\) and \(4.37 \mathrm{mg} \mathrm{H}_{2} \mathrm{O} .\) The molar mass of the compound is \(176.1 \mathrm{g} / \mathrm{mol} .\) What are the empirical and molecular formulas of the compound?

Short Answer

Expert verified
The empirical formula of the compound is C鈧咹鈧圤鈧佲倖, and the molecular formula is C鈧僅鈧凮鈧�.

Step by step solution

01

Calculate the moles of carbon and hydrogen

First, let's convert the mass of CO2 and H2O to moles. To do this, we need the molar masses of CO2 and H2O. The molar mass of CO2 is 12.01 g/mol (for C) + 2 * 16.00 g/mol (for O) = 44.01 g/mol. The molar mass of H2O is 2 * 1.01 g/mol (for H) + 16.00 g/mol (for O) = 18.02 g/mol. Now, let's convert the mass of CO2 and H2O to moles: - Moles of CO2 = (16.01 mg CO2) * (1 g / 1000 mg) * (1 mol CO2 / 44.01 g CO2) = \(3.636 \times10^{-4} mol\) - Moles of H2O = (4.37 mg H2O) * (1 g / 1000 mg) * (1 mol H2O / 18.02 g H2O) = \(2.427 \times10^{-4} mol\) Finally, let's determine the moles of carbon and hydrogen in the compound: - Moles of C = \(3.636 \times10^{-4} mol\) - Moles of H =\(2 \times 2.427 \times10^{-4} mol\) = \(4.854 \times10^{-4} mol\)
02

Calculate the moles of oxygen

We can find the moles of oxygen by using the initial moles of the compound. First, we need to convert the mass of compound to moles: - Moles of compound = (10.68 mg compound) * (1 g / 1000 mg) * (1 mol compound / 176.1 g compound) = \(6.063 \times 10^{-5} mol\) Since the combustion reaction is as follows: Compound + O鈧� 鈫� CO鈧� + H鈧侽 Total moles of O atoms in the compound = (total moles of O atoms in CO鈧�) + (total moles of O atoms in H鈧侽) Let's find the total moles of O atoms in CO鈧� and H鈧侽: - Moles of O in CO鈧� = \(2 \times 3.636 \times 10^{-4} mol\) = \(7.273 \times 10^{-4} mol\) - Moles of O in H鈧侽 = \(2.427 \times 10^{-4} mol\) Now, let's calculate the moles of O in the compound: - Moles of O = \(7.273 \times 10^{-4} mol + 2.427\times 10^{-4} mol\) = \(9.7 \times 10^{-4} mol\)
03

Determine the simplest whole-number ratio of elements (empirical formula)

Divide each number of moles by the smallest value to obtain the simplest whole-number ratio: \(C : \frac{3.636 \times10^{-4}}{6.063 \times 10^{-5}} = 6 ; H : \frac{4.854 \times10^{-4}}{6.063\times 10^{-5}} = 8 ; O : \frac{9.7 \times10^{-4}}{6.063\times 10^{-5}} = 16\) So, the empirical formula is C鈧咹鈧圤鈧佲倖.
04

Determine the molecular formula

First, calculate the empirical formula mass: C鈧咹鈧圤鈧佲倖: \(6(12.01)+8(1.01)+16(16.00) \approx 352.20\, g/mol \) Since the molar mass of the compound is about half of the empirical formula mass, the molecular formula should be half the empirical formula. Therefore, the molecular formula is C鈧僅鈧凮鈧�.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combustion Analysis
Combustion analysis is a technique used to determine the elemental composition of chemical compounds. It is particularly useful for organic compounds, which primarily contain carbon, hydrogen, and oxygen. During combustion analysis, a compound is burned in excess oxygen, producing carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)). By measuring these products, the amounts of carbon and hydrogen in the original compound can be determined.When performing combustion analysis:
  • The mass of \(\text{CO}_2\) is converted to moles to find the moles of carbon, since each mole of \(\text{CO}_2\) contains one mole of carbon.
  • Similarly, the mass of \(\text{H}_2\text{O}\) is converted to moles to find the hydrogen, as each mole of water contains two moles of hydrogen.
This process allows for the accurate determination of the compound's carbon and hydrogen content. Knowing the amounts of these elements is crucial for figuring out the empirical formula of the compound.
Stoichiometry
Stoichiometry is a core concept in chemistry that deals with the quantitative relationships between the elements in a chemical reaction. It plays a key role in predicting the amounts of products and reactants involved. Through stoichiometry, chemists can:
  • Balance chemical equations to ensure mass conservation.
  • Calculate the amounts of substances consumed or produced in a reaction.
  • Understand the mole ratio between different reactants and products.
In combustion analysis, stoichiometry allows us to ensure that the calculated moles of elements (like C, H, and O) accurately correspond to their proportional amounts in the compound. This accurate comparison is essential for determining empirical and molecular formulas.
Mole Calculations
Mole calculations are fundamental to understanding chemical reactions and the composition of compounds. The mole is a standard unit in chemistry that represents \(6.022 \times 10^{23}\) entities (atoms, molecules, or ions), often referred to as Avogadro's number.In the context of combustion analysis:
  • We convert the given mass of a substance into moles using its molar mass as a conversion factor.
  • This conversion allows for direct comparisons of the number of atoms and molecules involved, rather than their masses.
  • Mole calculations help determine the number of moles of carbon and hydrogen from \(\text{CO}_2\) and \(\text{H}_2\text{O}\), respectively.
Accurate mole calculations are crucial, as they form the basis for deriving the empirical formula by providing the correct ratios of the different elements in the compound.
Chemical Formula Determination
Determining the chemical formula of a compound involves figuring out two key formulas: the empirical formula and the molecular formula.
  • The empirical formula represents the simplest whole-number ratio of the elements in a compound. It is derived from combustion data and indicates the relative proportions of the constituent elements.
  • The molecular formula may be a multiple of the empirical formula and gives the actual number of each type of atom in a molecule of the compound.
To find the empirical formula, we divide the moles of each element by the smallest number of moles calculated, obtaining whole number ratios. For the molecular formula, compare the empirical formula mass to the compound's actual molar mass. If necessary, adjust the formula accordingly. This structured approach ensures an accurate representation of the compound鈥檚 composition, providing essential insights into its chemical nature.

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Most popular questions from this chapter

DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction: $$2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{C}_{2} \mathrm{HOCl}_{3} \longrightarrow \mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}+\mathrm{H}_{2} \mathrm{O}$$ In a government lab, 1142 g of chlorobenzene is reacted with 485 g of chloral. a. What mass of DDT is formed, assuming \(100 \%\) yield? b. Which reactant is limiting? Which is in excess? c. What mass of the excess reactant is left over? d. If the actual yield of DDT is \(200.0 \mathrm{g},\) what is the percent yield?

A common demonstration in chemistry courses involves adding a tiny speck of manganese(IV) oxide to a concentrated hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) solution. Hydrogen peroxide decomposes quite spectacularly under these conditions to produce oxygen gas and steam (water vapor). Manganese(IV) oxide is a catalyst for the decomposition of hydrogen peroxide and is not consumed in the reaction. Write the balanced equation for the decomposition reaction of hydrogen peroxide.

Balance each of the following chemical equations. a. \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) b. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{PCl}_{5}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{HCl}(g)\) e. \(\mathrm{CaO}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)\) f. \(\operatorname{MoS}_{2}(s)+\mathrm{O}_{2}(g) \rightarrow \operatorname{MoO}_{3}(s)+\mathrm{SO}_{2}(g)\) g. \(\mathrm{FeCO}_{3}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)\)

Hydrogen peroxide is used as a cleansing agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes on contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams on contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide: $$\mathrm{BaO}_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{BaCl}_{2}(a q)$$ What mass of hydrogen peroxide should result when \(1.50 \mathrm{g}\) barium peroxide is treated with \(25.0 \mathrm{mL}\) hydrochloric acid solution containing \(0.0272 \mathrm{g}\) HCl per mL? What mass of which reagent is left unreacted?

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. \(\mathrm{A}\) \(752-\mathrm{g}\) sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{g}\) of pure iron by the following reaction: $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(g)$$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.
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